Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$
What you can do is treat it like this:
$\displaystyle [6 - (13x - 6x^2)]^5$
Then apply the formula to find the first three terms.
$\displaystyle (a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...$
Taking a = 6
b = - (13x - 6x^2)
Hello, Punch!
This is a messy one . . . even with a good procedure.
$\displaystyle \text{Find the first 3 terms of: }\;(6-13x+6x^2)^5$
It is evident that the first three terms are:
. . . . $\displaystyle \text{constant},\;x\text{-term},\;x^2\text{-term}$
$\displaystyle \text{We have: }\:(a + b + c)^5\quad \text{where: }\:\begin{Bmatrix}a &=& 6 \\ b &=& \text{-}13x \\ c &=& 6x^2 \end{Bmatrix}$
The "Trinomial Expansion" forms an array:
$\displaystyle \begin{array}{cccccccccccc}&&& {5\choose5,0,0}a^5 \\ && {5\choose4,1,0}a^4b && {5\choose4,0,1}a^4c \\ & {5\choose3,2,0}a^3b^2 && {5\choose3,1,1}a^3bc && {5\choose3,0,2}a^3c^2 \\ {5\choose2,3,0}a^2b^3 && {5\choose2,2,1}a^2b^2c && {5\choose2,1,2}a^2bc^2 && {5\choose2,0,3}a^2c^3 \\ \vdots && \vdots && \vdots && \vdots\end{array}$
We want only the first four terms (reading from the top).
. . [The rest contain $\displaystyle x^3$ or higher powers.]
. . $\displaystyle {5\choose5,0,0}(6^5) + {5\choose4,1,0}(6^4)(\text{-}13x) + {5\choose4,0,1}(6^4)(6x^2) + {5\choose3,2,0}(6^3)(\text{-}13x)^2 $
. . $\displaystyle =\;1(7776) + 5(1296)(\text{-}13x) + 5(1296)(6x^2) + 10(216)(169x^2)$
. . $\displaystyle =\;7,\!776 - 84,\!240x + 38,\!880x^2 + 365,\!040x^2$
. . $\displaystyle =\;7,\!776 - 84,\!240x + 403,\!920x^2$
But check my work . . . please!
or look at it like this
$\displaystyle (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$
and so when doing u'll get
$\displaystyle T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5$
$\displaystyle T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4$
$\displaystyle T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]=-360x^7+1440x^6-2160x^5+1440x^4-360x^3$
u can go further... but u get the idea
or am i wrong here ?
hmmmm
probably lost in translation, but if u look at $\displaystyle (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$ then first one should be :
$\displaystyle T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0$
and so on...
but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that
Edit: Sorry, but now i'm little a bit of confused